87 CBE341 Lecture 11 (10/6) – Constitutive equations for a flowing fluid What have we derived thus far in the course? Till now, I focused on deriving conservation equations by using the principle of mass conservation and Newton’s law of linear momentum balance. Let us have a recap of the equations. I first formulated the integral mass balance equation. See equation (36) of my notes. Then I deduced the differential form of this balance, commonly referred to as the continuity equation. See equation (43) (for a general case). For the special case of an incompressible fluid, we obtain equation (46). Then I started with Newton’s law for linear momentum balance. I derived the integral form of this balance applicable to an arbitrary control volume. See equation (56) of my notes. Then I introduced the notion of a deviatoric stress tensor, see equations (117) and (118). Finally, I converted equation (56) to a differential form, referred to as the Cauchy’s equations of motion; see equations (129) and (136). We will now discuss the additional information needed before we can use the continuity equation and the Cauchy’s momentum balance equations to find detailed solutions for the velocity profiles in some flow problems. The continuity equation and the Cauchy’s momentum balance equations are reproduced below for convenience: ()0t¶r+ Ñ× r =¶v (43) ()() Pt¶r+ Ñ× r = r-Ñ+ Ñ׶vvv gt (129)88 Or equivalently, Pt¶r + r×Ñ = r-Ñ+ Ñ׶vvv gt (136) Many flow problems involve only modest pressure drops and the volume change is small even for gases. Hence, even gas flow problems can be treated as incompressible fluid flow problems with little loss of accuracy. For incompressible fluids, equation (43) simplifies to 0Ñ× =v (46) [Aside: If we consider examples such as flow of a gas through a very long pipe, with correspondingly large pressure drops and appreciable volume changes, then we retain equation (43). Even in these problems, the effect of compressibility in the momentum balance equations (namely 129 or 136) turns out to be weak, provided we are not talking about flows involving high Mach numbers (close to unity or larger). In chemical engineering, we rarely deal with such problems and so we can ignore the effect of compressibility in equations (129) and (136).] In this course, where we want to illustrate the use of balance equations to solve flow problems, we will only consider essentially incompressible flows. So, we will discuss the constitutive equations for the stress tensor only for the case of incompressible fluids. It was already mentioned that the viscous stress tensor is symmetric. Note that the viscous stress depends on the velocity gradient. For example, consider the so-called plane Couette flow.89 According to our notations, the viscous traction exerted by the upper plate on the fluid (sandwiched between the two plates) is given by = ×tnt, where n is the unit normal pointing from the entity on which the traction is being exerted (in this case, the fluid) towards the agent exerting the traction (in this case the upper plate). That is, Viscous traction = yyxxyyyyzz= ××t+ t + ttn e e e et = t = One can expect intuitively that, in order to sustain this motion, the upper plate must exert a force in the positive x-direction on the fluid. This force is simply the product of the area of contact between the fluid and the upper plate and yx xt e of the viscous traction. This component of the shear stress, yxt, depends on the velocity gradient xdvdy. For Newtonian fluids, yxt, depends on both xdvdy and ydvdx (see equation 156 at the end of these notes), however, in this example vy = 0. Based on this, one can anticipate that in a more general flow problem where the flow may not be one-dimensional, the stress tensor will depend on all possible velocity gradients, yyyxxx zz zdv dv dvdv dv dv dv dv dv,,,,,,,&.dx dy dz dx dy dz dx dy dz This is equivalent to saying that the stress tensor depends on the velocity gradient. =tt(Ñv) where ii=1,3jj=1,3v= xij¶Ñ¶åvee.90 In general, however, the velocity gradient, Ñv, is not symmetric. For example, in the plane Couette flow problem, 8 of the nine components of the velocity gradient Ñv are zero and xdvdy is the only non-zero element. However, the stress tensor has to be symmetric! This dilemma is resolved mathematically by demanding that the viscous stress tensor depends on the symmetric part of the velocity gradient tensor. Recall that the symmetric part of a two-dimensional square matrix, A = {aij} is simply {(aij + aji)/2}. In the same manner the symmetric part of the velocity gradient tensor is simply ( ) ( ) ( )jisymmetrici=1,3jij=1,3vv111= 2x x 2 2Tijæö¶¶Ñ +=Ñ + Ñç÷ç÷¶¶èøåveevv (140) where the superscript T denotes the transpose. What is the physical meaning of this requirement? Consider a fluid element as shown on the left below (in figure a). Let us suppose that after a little time the fluid element has translated and also changed the shape to assume the form shown on the right side. Translation of a fluid element does not produce any stress (absence of deformation). The change of shape clearly implies that the various corners of the original fluid element have moved at different rates, i.e. the velocities of the various corners of the fluid element are different. That is, there are some non-zero entries in the velocity gradient tensor! Now the change in the configuration of the fluid element (where I am not interested in the translation any more) consists of two parts: a deformation and a rotation.91 Figure a This is illustrated below, where we first deform the shape (see figure b) Figure b and subsequently rotate the element to get the final configuration (see figure c).92 Figure c Physically, the rotation of a fluid element does not produce any stress. So the operation in figure (c) should not play any role in computing the viscous stress, and only the first step matters. This is precisely what is achieved by taking the symmetric part of the velocity